Find the number of ways in which two Americans, two British, one Chinese, one Dutuch, and one Egyptian can sit on a round table so that persons of the same nationality are separated.

Explanation

There are seven person.
Total number of arrangements in circle without any restrictions
n(U)=6!
Let n(A) be the number of arrangements in which two Americans `A_(1) " and " A_(2)` are together.
`therefore n(A)=5! 2! " " ("considering" A_(1)A_(2)` as one unit)
Let n(B) be the number of arrangements in which two British `B_(1) " and " B_(2)` are together.
`therefore n_(B)=5!2! " " ("considering" B_(1)B_(2)` as one unit)
`therefore n(A cap B)`=number of arrangements in which two Americans are together and two British are together.
=4! 2! 2!
Now we want number of arrangements in which persons of the same nationality are separated.
i.e., `n(A' cap B')=n(U)-n(A cup B)`
=n(U)-[n(A)+n(B)-n(A cap B)]`
`=6!-[5!2!+5!2!-4! 2! 2!]`
=720-[240+240-96]
=720-384
=336